Bounded, Closed, and Compact Sets

Bounded, Closed, and Compact Sets Definition Let D be a subset of R n. Then D is said to be bounded if there is a number M > 0 such that x < M for all x D. D is closed if it contains all the boundary points. If D is both closed and bounded then it is said to be compact. Quang T. Bach Math 20C November 17, 2017 1 / 14

Question 1. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + y 2 + z 2 < 1} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, 2017 2 / 14

Question 2. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + z 2 < 1} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, 2017 3 / 14

Question 3. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + y 2 + z 2 = 2} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, 2017 4 / 14

Question 4. Consider the following subset of R 3 : U = {(x, y, z) : x + y + z = 2} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, 2017 5 / 14

Lagrange s Multiplier Method for Finding Extrema Let f, g : R n R be C 1 functions. Suppose we want to find the maximum and minimum of f (x) restricted to the constraint g(x) = c, for some constant c. Theorem If x 0 is an extrema of f (x) subject to the constraint g(x) = c and if g(x 0 ) 0 then f (x 0 ) = λ g(x 0 ) for some scalar λ. Quang T. Bach Math 20C November 17, 2017 6 / 14

Some Remarks This theorem does not guarantee the existence of the extrema of f (x) subject to the constraint g(x) = c. It simply says that if there is an extrema, then here is how to find possible candidates for it. If there is no x 0 that satisfies f (x 0 ) = λ g(x 0 ) and g(x 0 ) = c, then f does not have an extreme subject to the constraint g(x) = c. The solution to the equation f (x 0 ) = λ g(x 0 ) is called the critical point of f subject to the constraint g(x) = c. Quang T. Bach Math 20C November 17, 2017 7 / 14

The Step Lagrange s Multiplier Method Let f, g : R n R be C 1 functions. Suppose we want to find the maximum and minimum of f (x) restricted to the constraint g(x) = c, for some constant c. We shall find the potential candidates for the extrema by solving the following system: f (x 1,..., x n ) = λ g (x 1,..., x n ) x 1 x 1 f (x 1,..., x n ) = λ g (x 1,..., x n ) x 2 x 2. f (x 1,..., x n ) = λ g (x 1,..., x n ) x n x n g(x 1,..., x n ) = c Quang T. Bach Math 20C November 17, 2017 8 / 14

Example Example Find the point on the circle x 2 + y 2 = 1 that is closest to the point (4, 2). The distance between any point (x, y) and (4, 2) is given by d(x, y) = (x 4) 2 + (y + 2) 2. Notice that d(x, y) is minimal if and only if (d(x, y)) 2 is minimal, so we shall consider the objective function subject to the constraint f (x, y) = (x 4) 2 + (y + 2) 2 g(x, y) = x 2 + y 2 = 1 Quang T. Bach Math 20C November 17, 2017 9 / 14

Example The Lagrange s multiplier conditions are: f (x, y) = λ g(x, y) g(x, y) = 1 2(x 4) = λ(2x) 2(y + 2) = λ(2y) x 2 + y 2 = 1 x 4 = xλ y + 2 = yλ x 2 + y 2 = 1 From the 1st eqn, we observe that x 0 which then leads to λ = x 4. x From the 2nd eqn, we observe that y 0 which then leads to λ = y + 2. y So x 4 = λ = y + 2 xy 4y = xy + 2x x = 2y. x y Now subs x = 2y to the last eqn: ( 2y) 2 + y 2 = 1 y = ±1/ 5. So x = 2y = 2/ 5. Quang T. Bach Math 20C November 17, 2017 9 / 14

Example We obtain two critical points ( 2 ) ( 1 2, and 5, 1 ). 5 5 5 In this case, since the constraint is a closed and bounded set, f must have a maximum and minimum. In fact, these extrema are given by the solutions above. Here, ( f 2 ) 1, = 29.9443 5 5 f ( 2 5, 1 5 ) = 12.0557 ( 2 5, ) 1 is the maximum of f (x, y) on the unit circle and 5 So ( 2 5, 1 ) is the minimum of f on the unit circle. 5 Quang T. Bach Math 20C November 17, 2017 9 / 14

Example Example A manufacturer s production is modeled by the Cobb-Douglas function P(L, K) = 100L 3/4 K 1/4 where L represents the units of labor and K represents the units of capital. Each labor unit costs $200 and each capital unit costs $250. The total expenses for labor and capital is $50,000. Find the maximum production level. The objective function is subject to the constraint P(L, K) = 100L 3/4 K 1/4 g(l, K) = 200L + 250K = 50000 The Lagrange s multiplier conditions are: 75L 1/4 K 1/4 = 200λ Quang T. Bach Math 20C November 17, 2017 10 / 14

Example From the first two equations we obtain 3 8 L 1/4 K 1/4 = λ = 1 10 L3/4 K 3/4 15 4 K = L. ( ) 15 Now from the last equation: 200 4 K + 250K = 50000 K = 50. So L = 187.5. Quang T. Bach Math 20C November 17, 2017 10 / 14

Example Here, we obtain a critical point (187.5, 50). The constraint 200L + 250K = 50000 represents a line segment which is closed and bounded so P(L, K) must have a max and a min. The extrema must occur at either the critical point or the end points. P(0, 200) = P(250, 0) = 0 P(187.5, 50) 13473.9 so (187.5, 50) is the maximum of P(L, K) subject to the constraint 200L + 250K = 50000. The maximum production level is 13473.9 Quang T. Bach Math 20C November 17, 2017 10 / 14

Example - HW6 Problem 4a Example Write the number 120 as a sum of three numbers so that the sum of the products between any pair of two numbers is a maximum. Solution: Let x, y, z denote the three numbers. The objective function is subject to the constraint f (x, y, z) = xy + xz + yz g(x, y, z) = x + y + z = 120. Quang T. Bach Math 20C November 17, 2017 11 / 14

Example - HW6 Problem 4a The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 120 y + z = λ x + z = λ x + y = λ x + y + z = 120 The first three equations give x = y = z. Substitute this into the fourth equation to obtain x = y = z = 40. So (x, y, z) = (40, 40, 40) is the only critical point obtained from the method. Here, f (40, 40, 40) = 4800. Quang T. Bach Math 20C November 17, 2017 11 / 14

Example - HW6 Problem 4b Example Find three positive numbers x, y, and z whose sum is 10 such that x 2 y 2 z is a maximum. Solution: Let 0 < x, y, z < 10 denote the three numbers. The objective function is f (x, y, z) = x 2 y 2 z subject to the constraint g(x, y, z) = x + y + z = 10. Quang T. Bach Math 20C November 17, 2017 12 / 14

Example - HW6 Problem 4b The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 10 2xy 2 z = λ 2x 2 yz = λ x 2 y 2 = λ x + y + z = 10 The first three equations give x = y = 2z. Substitute this into the fourth equation to obtain x = y = 4 and z = 2. So (x, y, z) = (4, 4, 2) is the only critical point obtained from the method. Here, f (4, 4, 2) = 512. Quang T. Bach Math 20C November 17, 2017 12 / 14

Example - HW6 Problem 4c Example A wooden box is to be made with $120.00 worth of wood. The lid is to be made from wood that costs $2.00 per square inch and the rest of the box is to be made from wood that costs $2.50 per square inch. What is the biggest box that could be made? Solution: Let x, y, z > 0 denote the length, width, and height of the box. The objective function is subject to the constraint f (x, y, z) = xyz g(x, y, z) = 4.5xy + 5xz + 5yz = 120. Quang T. Bach Math 20C November 17, 2017 13 / 14

Example - HW6 Problem 4c The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 120 xz = λ(4.5x + 5z) yz = λ(4.5y + 5z) xy = λ(5x + 5y) 4.5xy + 5yz + 5xz = 120 Recall that x, y, z > 0 so from the first two equations, we obtain: xz 4.5x + 5z = λ = yz 4.5 + 5z x = y. With x = y, now from the third equation: x 2 = 10λx λ = x/10. Now subs λ = x/10 back to the first equation to obtain: xz = x 9x (4.5x + 5z) z = 10 10. Quang T. Bach Math 20C November 17, 2017 13 / 14

Example - HW6 Problem 4c Finally, we subs x = y, z = 9x/10 into the last equation to obtain: 4.5xy + 5yz + 5xz = 120 13.5x 2 = 120 x 2 = 80 9 x = 4 5 3. The solution ( obtain from the Lagrange s multiplier method is 4 5 (x, y, z) = 3, 4 5 3, 6 ) ( 5 4 5 and f 5 3, 4 5 3, 6 ) 5 = 32 5. 5 3 Quang T. Bach Math 20C November 17, 2017 13 / 14

Some Final Remarks In all 3 examples from HW6 above, I never claimed that the solution obtained from the Lagrange s multiplier method is the maximum of f (x, y, z) subject to the constraint g(x, y, z) = c. There is no guarantee that the solutions obtained from Lagrange s multiplier method is the extrema of the function. In fact, in all three cases above, the constraint region is not closed and bounded, so there is no guarantee that a global max/min even exists! Even thought the solutions we obtained here are all maximum of f, further analysis is required to obtain this conclusion. Quang T. Bach Math 20C November 17, 2017 14 / 14